I have been thinking about the physical meaning behind voltage and have come up with a few questions.

I think of voltage as being the amount of energy per coulomb of charge (J/C). Having a voltage across points A and B, say, means that the potential energy at point A is different (lets just say higher) than the potential energy at point B. I like to think of electric potential energy as being analogous to that of its gravitational counterpart; lifting up a mass up a distance gives it potential energy, lifting it higher gives it more potential energy etc.

If I have some charge, say 10C of +ve charge, and I grab a unit negative charge and hold it in place a certain distance from my positive bunch, we say that the negative charge has potential energy (since it WANTS to move towards the positive charges). If I provide a medium or a path for this negative charge to move (say a wire) to the less potential state (closer to the +ve charge) then this results in a current , which I like to think of as the number of charges that pass through a cross section of this wire to the other point B of lower potential.

So far so good? Please correct me if i'm wrong.

Now, my question is as follows:
How does a battery cell (galvanic cell) generate a point of higher potential A with respect to point B? My understanding is that the chemical reactions give rise to the "generation" of ELECTRONS (negative charge) but how does that cause one end to be of higher potential than the other? (ie: how does this link back to the explanation of voltage as being the fixing of a charge at a distance away from an opposing charge etc..? or the analogy to gravitational potential?)

Keep in mind a battery does not generate electrons it just moves them from the positive plate to the negative plate. Look at how a vandegraf generator does it: puts the charge on a conveigher (sp) belt.

Does the accumulation of charge on a plate inside the battery give each electron more energy (hence creating a larger voltage J/C) ? Is it because of the repulsion between the charges that gives rise to this increased potential energy?

You could start with quantum mechanics, which explains that the energy level is sometimes lower to have an ionized atom or molecule that has filled electron shells rather than what would otherwise be expected - a non-ionized atom or molecule with a missing spot in its outer electron shell or an additional electron all by itself outside a filled shell.

The same thing is also true of more complex molecules - sometimes a particular compound will settle into a lower energy state even if it involves gaining or losing an electron or two - the key is that the resulting compound is at a lower energy state, and just like a ball rolling downhill, molecules will tend to form into lower energy states.

Take a lead-acid battery. It's got a negative plate of lead, Pb, and a positive plate of lead oxide, PbO2. Between them is a sulfuric acid solution, H2SO4 and water, H2O. Not to go into every little detail, but the key point is that it's a lower energy state for lead molecules to lose two electrons and form lead sulfate PbSO4 at the negative plate, and for lead oxide to absorb two extra electrons and go from PbO2 to PbSO4 (and the O2 combines with the leftover hydrogen from the H2SO4 and forms 2H2O).

Since the reaction is moving molecules to a lower energy state, energy is released in
undergoing the reaction; that energy is absorbed in forcing the charges onto the already partially charged terminals. So the charges that have been forced onto the terminals are like a stretched spring, or a ball at the top of a hill; there is energy available by allowing the charges to recombine via connecting a load across the battery terminals.

As the battery terminals build up charge, the voltage rises, until eventually the additional force needed to add additional charge to the terminals just balances the force exuded by the chemically reacting molecules, and the chemical reactions temporarily stop.

But in a charged battery, there is still plenty of unreacted lead and lead oxide; these
represent the potential energy in the battery. They "want" to go to a lower energy state, but as long as the terminals are charged to the natural battery voltage, the molecules are unable to move the electrons needed to allow themselves to react into the lower energy state.

If charge is drawn off both terminals by an external load, then the battery is once again unbalanced and more internal lead and lead oxide will turn into lead sulfate, forcing charge onto the terminals in the process.

Voltage is just the measure of the force it takes to move charges to the terminals. The more unbalanced charges between the terminals, the more voltage will be measured. The overall potential energy in a battery is a different thing; it's more of a measure of how much unreacted compounds are waiting to react, along with how hard they do react, which would correspond to how much voltage will accumulate between the terminals.

[QUOTEDoes the accumulation of charge on a plate inside the battery give each electron more energy (hence creating a larger voltage J/C) ?][/QUOTE]

More energy than what? The accumulation of charge anywhere creates a voltage, because it took energy to accumulate the charge. Voltage is not potential energy, voltage is the energy density of the charge. See the links below.